MODEL: DLS

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DLS SPECIFICATIONS

ES-6515 Rev.18 Page 4/10

unit (0.5 sec.)

[example] Modem type /M5A (1200 bps)

2 × 12 ÷ 1200 + 0.5 = 0.52 (sec.)

3. Total Transmission Time per Unit

The above (1) transmission time and (2) & (3) Start-/End-of transmission time must be added to calculate the total transmission

time required by one DLS unit.

Total transmission time per unit = (start-of-transmission time) + (transmission time) + (end-of-transmission time)

[example] Station A: S1 and A1 units

Station B: S1 and C1 units

Modem type /M5A (1200 bps)

•Station A transmission time

0.12 + (1.18 + 0.8) + 0.52 = 2.62 (sec.)

•Station B transmission time

0.12 + (1.18 + 0) + 0.52 = 1.82 (sec.)

4. Overall Transmission Cycle

The overall transmission cycle is determined as the time required by one DLS unit starting transmission before the next

transmission.

Overall transmission cycle = Station A total transmission time + Station B total transmission time

[example] Station A: S1 and A1 units

Station B: S1 and C1 units

Modem type /M5A (1200 bps)

2.62 + 1.82 = 4.44 (sec.)

The time required for the input signals to be output at the output unit varies according to the exact moment of input.

•Minimum (Station A input to Station B output)

= Station A total transmission time per unit

= 2.62 (sec.)

•Maximum (Station A input ot Station B output)

= Overall transmission cycle

+ Station A total tansmission time per unit

= 7.06 (sec.)

Therefore, the time required for a Station A input to be output at the Station B varies between 2.62 and 7.06 seconds as far as

there is no transmission error.